Answer:
a) 0.76197086
b) -1.73406361
Explanation:
a)
Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29)
P(-1.29 < t < 1.29) would be the area under the t distribution curve with 7 degrees of freedom between -1.29 and 1.29, that is in the interval (-1.29, 1.29).
This can be done the old style by looking up in a table or by using the technology with a spreadsheet.
In Excel, the function TDIST(x,n,2) with x>0 gives the area outside the interval (-x, x) of the t distribution with n degrees of freedom.
So TDIST(1.29,7,2) gives the area outside (-1.29, 1.29).
If we subtract this value from 1 we get the desired result
Hence
P(-1.29 < t < 1.29) = 1 - TDIST(1.29,7,2) = 1 - 0.23802914 = 0.76197086
In OpenOffice Calc, the function is the same replacing “,” with “;”
That is
P(-1.29 < t < 1.29) = 1 - TDIST(1.29;7;2) = 0.76197086
b)
Consider a t distribution with 18 degrees of freedom. Find the value of c such that P(t≤ c) = 0.05
We are looking for a point c such that the area of the t distribution with 18 degrees of freedom to the left of c is 0.05
In Excel, the inverse function of TDIST is TINV.
TINV(p*2,n) with p>0 gives the point c such that the area of the t distribution with n degrees of freedom to the right of c is p.
Since the t distribution is symmetric with respect to 0, -c would be a point such that the area to the left of -c is p.
So we want to compute in Excel
-TINV(0.05*2,18) = -1.73406361
In OpenOffice Calc
-TINV(0.05*2;18) = -1.73406361